Question

The average loan amount issued by a small short-term lender is $1080 with a standard deviation of $184. Determine the probabilities, assuming that the population data is normally distributed.

a) What is the probability that the lender issues more than $1150 to a random borrower?

b) What is the probability that the lender issues at most $1150, on average, to a random sample of 30 borrowers?

c) What is the probability that the lender issues between $1150 and $1175, on average, to a random sample of 30 borrowers?

Answer 1

Solution :

Given that,

mean = = 1080

standard deviation = = 184

a ) P (x > 1150 )

= 1 - P (x < 1150 )

= 1 - P ( x -  / ) < ( 1150 - 1080 / 184)

= 1 - P ( z < 70 / 184 )

= 1 - P ( z < 0.83)

Using z table

= 1 - 0.7967

= 0.7967

Probability = 0.7967

b )n = 30

= 1080

₌ / n =18410 =33.5936

p ( 1150 )

P ( - /)   (1150 - 1080 /33.5936)

P( z   70 / 33.5936 )

P ( z 2.08 )   

Using z table

= 0.9812

Probability = 0.9812

c ) n =30

= 1080

₌ / n =184 30 =33.5936

P (1150 << 1175 )

P ( 1150- 1080 / 33.5936) < ( - / ) < ( 1175 - 1080 / 33.5936)

P ( 70 / 33.5936 < z < 90 / 33.5936 )

P (2.08 < z < 2.83)

P ( z < 2.83 ) - P ( z < 2.08)

Using z table

= 0.9977 - 0.9812

= 0.0165

Probability = 0.0165